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In my assignment on quantum mechanics, I have encountered a sum $$\sum_{n\neq 1}\langle nlm|z|100\rangle$$ for hydrogen states $|nlm\rangle$. From the selection rule, we know that $l$ can only be $1$, $m$ can only be $0$. The problem will be reduced to $$\sum_{n\neq 1}\langle n10|z|100\rangle$$ but still, in the coordinate basis, the inner product integral is difficult to evaluate for large $n$ (not even talk about the sum). So, is there any good idea on how to evaluate this sum?

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  • $\begingroup$ Are you sure that you need to evaluate the infinite sum? It seems like a Herculean task for an assignment. $\endgroup$ Oct 24 '16 at 2:20
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Since $z = r\cos\theta$, the angular integral can be evaluated and the remaining difficulty is in the matrix elements $\langle nl|r|10\rangle$.

The paper (unfortunately paywalled)

presents various recurrence relations for matrix elements of the operator $r$. In particular equation (5.7) of the paper is

$$C^{-1}_{n,l-1} 2l \langle n' l |r|n,l-1 \rangle= C^{-1}_{n',l} (2l+1) \langle n', l+1|r|nl\rangle + C^{-1}_{nl} \langle n' l |r|n, l+1 \rangle \tag{5.7}$$ where $$C_{nl} = [1/(l+1)^2 - 1/n^2]^{-1/2} \tag{2.17}.$$

Even so, the infinite sum certainly seems intractable.

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